Explanation

(a) $W_{AB} = O;W_{BC} = 2P_{0}V_{0};W_{CD} = 0;W_{DA} = - P_{0}V_{0}$
So total work done $= P_{0}V_{0}$
From $A$ and $B$, heat given to the gas
$= nC_{v}\mathrm{\Delta}T = n\frac{3}{2}R\mathrm{\Delta}T = \frac{3}{2}V_{0}\mathrm{\Delta}P = \frac{3}{2}P_{0}V_{0}$
From $B$ to $C$, heat given to the system
$= nC_{p}\mathrm{\Delta}T = n\left( \frac{5}{2}R \right)\mathrm{\Delta}T = \frac{5}{2}\left( 2P_{0} \right)\mathrm{\Delta}V = 5P_{0}V_{0}$
From $C$ to $D$ and $D$ to $A$, heat is rejected
Efficiency, $\eta = \frac{work\ done\ by\ gas}{heat\ given\ to\ the\ gas} \times 100$
$\eta = \frac{P_{0}V_{0}}{\frac{3}{2}P_{0}V_{0} + 5P_{0}V_{0}} = 15.4\%$